When the tank was completely full, outlets X and Z were opened together for *H* hours. If outlet Z alone had been opened, it would have taken *H + 10* hours to empty the same quantity of water from the tank. Given that outlet X alone empties the tank in 34 hours and outlet Z alone empties the tank in 30 hours, what is the value of *H*?

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**Solution:**

1. Let the time for which outlets X and Z run together be *H* hours.

2. Outlet X empties the tank in 34 hours, so its rate is $\frac{1}{34}$ tank per hour.

3. Outlet Z empties the tank in 30 hours, so its rate is $\frac{1}{30}$ tank per hour.

4. When both outlets X and Z work together for *H* hours, the quantity emptied is:
   $$H 	imes \left(\frac{1}{34} + \frac{1}{30}ight)$$

5. If outlet Z alone empties the same quantity, it takes *H + 10* hours, so:
   $$\left(H + 10ight) 	imes \frac{1}{30}$$

6. Equate the two expressions for the quantity emptied:
   $$H \left(\frac{1}{34} + \frac{1}{30}ight) = \frac{H + 10}{30}$$

7. Simplify the left side:
   $$H \left(\frac{30 + 34}{34 	imes 30}ight) = \frac{H + 10}{30}$$

$$H 	imes \frac{64}{1020} = \frac{H + 10}{30}$$

8. Multiply both sides by 1020 to clear denominators:
   $$64H = 34(H + 10)$$

9. Expand the right side:
   $$64H = 34H + 340$$

10. Bring terms involving *H* to one side:
    $$64H - 34H = 340$$

$$30H = 340$$

11. Solve for *H*:
    $$H = \frac{340}{30} = \frac{34}{3} = 11 \frac{1}{3} 	ext{ hours}$$

**Answer:** $11 \frac{1}{3}$ hours

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